In the process of nuclear fission of 1 g of uranium, the mass lost is 0.90 mg. The efficiency of fission reactor of power house is 20 %. To obtain 400 MW power from the power house, how much uranium (in gram) is required per hour?

Question

In the process of nuclear fission of 1 g of uranium, the mass lost is 0.90 mg. The efficiency of fission reactor of power house is 20 %. To obtain 400 MW power from the power house, how much uranium (in gram) is required per hour? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Power =400 MW ∴ Energy obtained per hour =400 × 106 × 3600 =144 × 1010 J Input energy of power house =(144 × 1010 × 10/20)=72 × 1011 J Now, Δ m c2=72 × 1011 Δ m=(72 × 1011/9 × 1016) kg =0.08 g As mass lost in 1 g uranium is 0.9 × 10-3 g, ∴ Required mass =0.08 /(0.9 × 10-3) =8.89 g

Q. In the process of nuclear fission of 1g of uranium, the mass lost is 0.90mg. The efficiency of fission reactor of power house is 20%. To obtain 400MW power from the power house, how much uranium (in gram) is required per hour?

Power =400MW ∴ Energy obtained per hour =400×106×3600 =144×1010J Input energy of power house =20144×1010×10​=72×1011J Now, Δmc2=72×1011 Δm=9×101672×1011​kg=0.08g As mass lost in 1g uranium is 0.9×10−3g, ∴ Required mass =0.08/(0.9×10−3) =8.89g

Q. In the process of nuclear fission of $1 g$ of uranium, the mass lost is $0.90 m g$ . The efficiency of fission reactor of power house is $20%$ . To obtain $400 M W$ power from the power house, how much uranium (in gram) is required per hour?